Steam at $100^\circ\text{C}$ is passed into $20\text{ g}$ of water at $10^\circ\text{C}$. When water acquires a temperature of $80^\circ\text{C}$, the mass of water present will be: [Take specific heat of water $= 1\text{ cal /g }^\circ\text{C}$ and latent heat of steam $= 540\text{ cal g}^{-1}$]: (2014)
Solution:
Heat gained by water $= 20 \times 1 \times (80 - 10) = 1400\text{ cal}$. Heat lost by $m$ grams of steam $= m \times 540 + m \times 1 \times (100 - 80) = 560m$. Equating them: $560m = 1400 \Rightarrow m = 2.5\text{ g}$. Total mass $= 20 + 2.5 = 22.5\text{ g}$.
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